Mathematics Asked by Tug Witt on December 12, 2020
From my understanding of (smooth) manifolds, all you need is an atlas to describe a manifold. However, if you have some atlas ?={($U_n$,$phi_n$)} with $n$ charts, we still haven’t defined our transition maps. My questions are:
You can simply define the transition maps, once the atlas is given.
There is a transition map which I shall denote $psi_{m,n}$ for every pair of indices $m,n$ having the property that $U_m cap U_n ne emptyset$.
The domain of $psi_{m,n}$ is the set $phi_m(U_m cap U_n) subset mathbb R^k$ (I'm assuming implicitly that $k$ is the dimension of the manifold).
The range (or codomain) of $psi_{m,n}$ is the set $phi_n(U_m cap U_n) subset mathbb R^k$.
And the formula for $psi_{m,n} : phi_m(U_m cap U_n) to phi_n(U_m cap U_n)$ is $$psi_{m,n}(p) = phi_n(phi^{-1}_m(p)), quad p in phi_m(U_m cap U_n) $$
Also, once all of this is written down, one can use the definition of a manifold together with the Invariance of Domain Theorem to prove that the domain and range of $phi_{m,n}$ are both open subsets of $mathbb R^k$, and one can show that $psi_{n,m}$ is an inverse map of $psi_{m,n}$, hence each transition map is a homeomorphism from its domain to its range.
And once that is done, you can now ask yourself questions that are aimed at determining whether your manifold is a $C^infty$ manifold, or a $C^2$ manifold, or a $C^1$ manifold or whatever smoothness property you want. Namely: Are the functions ${psi_{m,n}}$ all $C^infty$? or are they all $C^2$? or $C^1$?
Correct answer by Lee Mosher on December 12, 2020
Once you have the charts $phi_n$, the transition maps are determined, as $phi_mcircphi_n^{-1}$. (That uses my favorite convention for the direction of these maps; you might need to move the "inverse" if your convention is different.)
Answered by Andreas Blass on December 12, 2020
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